Chapter 9.1, Problem 65PPSE

Solution

a.

To find the number of boxes need at the next lower level.

The next lower level will have 25 boxes.

Given information:

Stacking of boxes that forms a square.

Calculation:

The first level has 1 box, second has 4 boxes, third has 9 boxes and fourth has 16 boxes.

The corresponding sequence is $1,4,9,16,\mathrm{...}$ .

The nth term of this sequence is $a_n=n^2$ .

The next lower level will have $a_5=5^2=25$ boxes.

b.

To find the number of boxes required to add three levels.

It requires 110 boxes to add three levels.

Given information:

Stacking of boxes that forms a square.

Calculation:

The number of boxes required to add three levels will be:

  $\begin{array}{c}{a}_5+a_6+a_7=5^2+6^2+7^2\\ =25+36+49\\ =110\end{array}$

So, it requires 110 boxes to add three levels.

c.

To find the number levels when there are 285 boxes used.

The number of levels will be 9.

Given information:

Stacking of boxes that forms a square.

Calculation:

The given sequence is square of natural numbers.

Recall that the sum of squares of n natural numbers is given by $\frac{n\left(n+1\right)\left(2n+1\right)}{6}$ .

If there are 285 boxes, then $\frac{n\left(n+1\right)\left(2n+1\right)}{6}=285$ .

Now, solve this equation.

  $\begin{array}{c}\frac{n\left(n+1\right)\left(2n+1\right)}{6}=285\\ \frac{n\left(n+1\right)\left(2n+1\right)}{6}\cdot 6=285\cdot 6\\ n\left(n+1\right)\left(2n+1\right)=1710\\ 2n^3+3n^2+n=1710\\ 2n^3+3n^2+n-1710=0\\ \left(n-9\right)\left(2n^2+21n+190\right)=0\\ n-9=0,2n^2+21n+190=0\end{array}$

The equation $2n^2+21n+190=0$ has imaginary solutions. The only real solution is $n=9$ .

Therefore, the number of levels will be 9.